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Vertical motion with air friction

  1. Apr 23, 2006 #1
    An object falling from some point in the air (near the surface) with air friction, R = kv.

    So,

    [tex] m\frac{dv}{dt} = mg - kv[/tex]

    Seperate variables for

    [tex]\int \frac{m}{mg - kv} \cdot dv = \int dt[/tex]

    and for the LHS I use integration by parts, so,

    [tex](m)(mgv - \frac{1}{2}kv^2) - \int mgv - \frac{1}{2}kv^2 \cdot dv = t[/tex]

    am on the right track here?
     
  2. jcsd
  3. Apr 23, 2006 #2

    nrqed

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    Your integration by parts is incorrect (I have no idea what you used for ''u'' and ''v'' but there is no way to get what you wrote).
    The integral over v is basically a log, that's all.
     
  4. Apr 23, 2006 #3
    Woops, I see what I did. Let me try that again.

    But it should take the form

    [tex](m)\int (mg - kv)^{-1} \cdot dv - \int (1)\left(\int (mg - kv)^{-1}\right) \cdot dv = t[/tex]

    correct?

    edit: missed your part about a log. I guess we agree?
     
    Last edited: Apr 23, 2006
  5. Apr 23, 2006 #4

    nrqed

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    ? The first term is the integral you started with! I am not sure where the second term could come from.

    You can do the integral in one step, without using integ by parts
     
  6. Apr 23, 2006 #5
    asdfakj I wasn't thinking of that m as a constant for some reason :\ I see now, thanks.
     
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